MATH 253 WORKSHEET 23
POLAR COORDINATES AND INTEGRATION
1.
Polar coordinates
(1) Let D = (x, y) | 1 ≤ x + y ≤ 2, x, y ≥ 0 .
(a) Express D in the form D = {(r, θ) | ??}
√
Solution: The rst condition reads 1 ≤ r2 ≤ 2 or 1 ≤ r ≤ 2. For the second condition, geometrically we see that 0 ≤ θ ≤ π2 . Algebraically, x, y ≥ 0 means both r cos θ ≥ 0
and r sin θ ≥ 0. Since r is non-negative this means sin θ, cos θ ≥ 0. The rst means 0 ≤
θ ≤ π , and the second means − π2 ≤ θ ≤ π2 , so we again get 0 ≤ θ ≤ π2 and the region is
√
(r, θ) | 1 ≤ r ≤ 2, 0 ≤ θ ≤ π2 .
(b) Try expressing D cos x2 + y 2 dA as an iterated integral, slicing the domain vertically.
Solution: For 0 ≤√x ≤ 1, the vertical slices begin at the inner circle and end on the outer
circle. For 1 ≤ x ≤ 2, the vertical slices begin at the x-axis and end on the outer circle. The
integral is therfore
x=1 y=√2−x2
x=√2 y=√2−x2
2
dx
x=0
2
√
y= 1−x2
dy cos x2 + y 2 +
dy cos x2 + y 2
dx
x=1
y=0
(c) Calculate D cos x2 + y 2 dA in polar coordinates.
Solution: This is
θ= π2 ! θ= π2 r=√2
r dr cos r2
dθ
θ=0
=
θ=0
r=1
r=√2
π 1
sin(r2 )
2 2
=
r=1
!
cos r
dθ
r=1
=
√
r= 2
2
r dr
π (sin 2 − sin 1)
.
4
(2) Find the volume of the solid lying above the xy -plane, below the paraboloid z = x2 + y 2 and inside
the cylinder (x − 1)2 + y 2 = 1.
(a) Find a region R in the plane and a function f (x, y) so that the volume
is
f (x, y) dA.
R
Solution: Points inside the cylnder have (x, y) belonging to R = (x, y) | (x − 1)2 + y 2 ≤ 1 .
Volume is then
(x2 + y 2 ) dx dy
z dA =
R
R
(b) Write R and f in polar coordinates.
Solution: Setting x = r cos θ, y = r sin θ, we have r2 cos2 θ − 2r cos θ + 1 + r2 sin2 θ ≤ 1.
Subtracting the 1 and using cos2 θ + sin2 θ = 1 this isequivalent to r2 ≤ 2r cos θ, and dividing by
r (which is always positive) thisis equivalent to r ≤ 2 cos θ. Since
the region is on the right of
the y axis, − π2 ≤ θ ≤ π2 , so R = (r, θ) | − π2 ≤ θ ≤ π2 , r ≤ 2 cos θ . x2 + y 2 = r2 so f (r, θ) = r2 .
(c) Evaluate the integral.
Solution: See solution to WS 24
Date
: 1/11/2013.
1